1. Prepared By: Shakil Raiman

2.  Newton’s law of gravitation states that the gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation  The gravitational force F between two point masses M and m separated by a distance r is given by  where G = universal gravitational constant

3.  Gravitational field strength at a point is the gravitational force per unit mass at that point. It is a vector and its S.I. unit is N kg -1.  By definition,  By Newton Law of Gravitation,  Combining, magnitude of  Therefore g = GM / r 2, where M = Mass of object “creating” the field  The acceleration due to gravity is same as the gravitational field strength.  On earth g=9.81 m/s 2

4.  Gravitational potential at a point is defined as the work done (by an external agent) in bringing a unit mass from infinity to that point (without changing its kinetic energy).   As the gravitational force on the mass is attractive, the work done by an ext agent in bringing unit mass from infinity to any point in the field will be negative work (as the force exerted by the ext agent is opposite in direction to the displacement to ensure that ΔKE = 0)

5.  Gravitational potential energy U of a mass m at a point in the gravitational field of another mass M, is the work done in bringing that mass m (NOT: unit mass) from infinity to that point.   Change in GPE, ΔU = mgh only if g is constant over the distance h; ( h<< radius of planet) otherwise, must use: ΔU = mφ f -mφ i

6. AspectsElectric FieldGravitational Field 1.Quantity interacting with or producing the fieldCharge QMass M 2.Definition of Field Strength Force per unit positive charge E = F / q Force per unit mass g = F / M 3.Force between two Point Charges or Masses Coulomb's Law: F e = Q 1 Q 2 / 4πε o r 2 Newton's Law of Gravitation: F g = GMm / r 2 4.Field Strength of isolated Point Charge or MassE = Q / 4πε o r 2 g = GM / r 2 5.Definition of Potential Work done in bringing a unit positive charge from infinity to the point; V = W /Q Work done in bringing a unit mass from infinity to the point; φ = W / M 6.Potential of isolated Point Charge or MassV = Q / 4πε o rφ = -GM / r 7.Change in Potential EnergyΔU = q ΔVΔU = m Δφ

7.  Escape Speed of a Satellite  By Conservation of Energy, Initial KE+Initial GPE=Final KE+Final GPE  (½mv E 2 )+(-GMm / r)= 0  Thus escape speed, v E = √(2GM / R)  Note : Escape speed of an object is independent of its mass  For a satellite in circular orbit, "the centripetal force is provided by the gravitational force“  Hence GMm / r 2 = mv 2 / r = mrω 2 = mr (2π / T) 2  A satellite does not move in the direction of the gravitational force (ie it stays in its circular orbit) because: the gravitational force exerted by the Earth on the satellite is just sufficient to cause the centripetal acceleration but not enough to also pull it down towards the Earth. This explains also why the Moon does not fall towards the Earth.

8.  Geostationary satellite is one which is always above a certain point on the Earth (as the Earth rotates about its axis.)  For a geostationary orbit: T = 24 hrs, orbital radius (& height) are fixed values from the centre of the Earth, and velocity w is also a fixed value; rotates fr west to east. However, the mass of the satellite is NOT a particular value & hence the ke, gpe, & the centripetal force are also not fixed values {ie their values depend on the mass of the geostationary satellite.}  A geostationary orbit must lie in the equatorial plane of the earth because it must accelerate in a plane where the centre of Earth lies since the net force exerted on the satellite is the Earth's gravitational force, which is directed towards the centre of Earth.

12. Prepared By: Shakil Raiman

13.  Red Giant:  A giant star with a relatively low surface temperature, so that it glows with a red colour.  White Dwarf:  The very hot remnant of a low-mass star.  Blue Supergiant:  Very hot, massive, luminous star with a short lifespan observed in young cosmic structures.

14.  Stars are much too far away from us to send probes to them, or even to send signals to in hope of detecting reflections. However, scientists have managed to determine an enormous amount of detailed information from even the faintest glows in the night sky.  The EM emissions from stars can tell us their temperature, chemical composition, speed of movement, approximate age, size and much more.

15.  Luminosity:  The rate at which a source such as a star radiates energy.  Black body:  A black body is a perfect emitter and absorber of radiation. A black body absorbs all wavelengths that fall upon it and radiates all wavelengths. The spectrum of wavelengths emitted depends only on its temperature.

16.  Stefan-Boltzmann law states that the total power output from a black body is proportional to its surface area and the fourth power of its temperature in kelvin.  L =  T 4 × surface area  where  is the Stefan-Boltzmann constant.  = 5.67 ×10 -8 Wm 2 K -4  For a sphere, L = 4  r 2  T 4

17.  Wien’s law describes the relationship between the peak output wavelength and the temperature of a star.

23.  Graph of luminosity of stars against their surface temperature. On this diagram stars appear in groups related to the age and size of the stars.  Main Sequence: It is the main group of stars on a Hertzsprung-Russell diagram. These are stars that are burning hydrogen. Our sun is a main sequence star.

24.  Protostar: The first stage in the lifecycle of a star. The accreting collection of gases which will become a star once enough material is collected and nuclear fusion begins.  Neutron Star: A dense ball of neutrons that remains at the core of a star after a supernova explosion.  Black Hole: A remnant of a massive star formed when its matter collapses to a point. The density of a black hole is so great that even light cannot escape from it.  Planetary Nebula: The hot gases and dust ejected from a low-mass star as it evolves into a white dwarf.  Black dwarf: The theoretical endpoint of the life cycle of a low-mass star.

26.  Once it has accreted about the mass of our Sun, a low-mass star will start to undergo nuclear fusion of hydrogen, converting into helium.  This stage stays for billions of years.  When it will run low on hydrogen fuel, but will have produced so much energy that it will expand slightly.  Our Sun is expected to undergo this change in another 4-5 billion years, when it will expand beyond the orbit of Venus.  Once most the hydrogen fuel is used, the star will start fusing helium nuclei. This complex process can cause an explosion forming planetary nebula.  As fuel decreases, outward pressure drops, star contract and heats up significantly and become white dwarf.

27.  If a protostar has more than four times the mass of our Sun, the star begins life as a blue supergiant. Nuclear fusion begins and the star enters a stable stage of life here heat pressure and gravity are in equilibrium.  Produce high temp (core temp is 40 million K but for Sun it is 15 million K)  So burns very quickly and further fusion possible to produce larger atoms. The fusion of helium can produce a variety of larger elements and so on and at the end mostly iron.  Then it can no longer undergo fusion and stops producing energy which happens more abruptly than in a low mass star.  Due to larger mass with high gravitational force it undergo incredible collapse and causes an explosion which is called a (type II) supernova.  This produce either a neutron star or a blackhole.

33.  Wish you all very good luck and excellent result.